Bayes_Theorem - conditional__Probability

probability

1. Bayes’ Theorem
Special Type of Conditional Probability

2. Recall- Conditional Probability
P(Y  T  C|S) will be used to calculate
P(S|Y  T  C)
P(Y  T  C|F) will be used to calculate
P(F|Y  T  C)
HOW?????
We will learn in the next lesson?
BAYES THEOREM

3. Definition of Partition
Let the events B
1
, B
2
, , B
n
be non-empty subsets
of a sample space S for an experiment. The B
i
’s
are a partition of S if the intersection of any two of
them is empty, and if their union is S. This may be
stated symbolically in the following way.
1. B
i  B
j = , unless i = j.
2. B
1
 B
2
   B
n
= S.

4. Partition Example
S
B
1 B
2
B
3

5. Example 1
Your retail business is considering holding
a sidewalk sale promotion next Saturday.
Past experience indicates that the
probability of a successful sale is 60%, if it
does not rain. This drops to 30% if it does
rain on Saturday. A phone call to the
weather bureau finds an estimated
probability of 20% for rain. What is the
probability that you have a successful
sale?

6. Example 1
Events
R- rains next Saturday
N -does not rain next Saturday.
A -sale is successful
U- sale is unsuccessful.
Given
P(A|N) = 0.6 and P(A|R) = 0.3.
P(R) = 0.2.
In addition we know R and N are complementary events
P(N)=1-P(R)=0.8
Our goal is to compute P(A).
)RN(
c


7. Using Venn diagram –Method1
Event A is the
disjoint union of
event R  A
&
event N  A
S=RN
R
N
A
P(A) = P(R  A) + P(N  A)

8. P(A)- Probability that you have a
Successful Sale
We need P(R  A) and P(N  A)
Recall from conditional probability
P(R  A)= P(R )* P(A|R)=0.2*0.3=0.06

Similarly
P(N  A)= P(N )* P(A|N)=0.8*0.6=0.48
Using P(A) = P(R  A) + P(N  A)
=0.06+0.48=0.54

9. Let us examine P(A|R)

Consider P(A|R)
The conditional
probability that sale is
successful given that it
rains
Using conditional
probability formula
)R(P
)AR(P
)R|A(P


S=RN
R
N
A

10. Tree Diagram-Method 2
Bayes’, Partitions
Saturday
R
N
A R  A 0.20.3 = 0.06
A N  A 0.80.6 = 0.48
U R  U 0.20.7 = 0.14
U N  U 0.80.4 = 0.32
0.2
0.8
0.7
0.3
0.6
0.4
Probability
Conditional
Probability
ProbabilityEvent
*Each Branch of the tree represents the intersection of two events
*The four branches represent Mutually Exclusive events
P(R ) P(A|R)
P(N ) P(A|N)

11. Method 2-Tree Diagram
Using P(A) = P(R  A) + P(N  A)
=0.06+0.48=0.54

12. Extension of Example1
Consider P(R|A)
The conditional probability that it rains given
that sale is successful
the How do we calculate?
Using conditional probability formula
)N(P)N|A(P)R(P)R|A(P
)R(P)R|A(P
)A(P
)AR(P
)A|R(P





80602030
2030
....
..


=
= 0.1111
*show slide 7

13. Example 2
In a recent New York Times article, it was
reported that light trucks, which include
SUV’s, pick-up trucks and minivans,
accounted for 40% of all personal vehicles on
the road in 2002. Assume the rest are cars.
Of every 100,000 car accidents, 20 involve a
fatality; of every 100,000 light truck accidents,
25 involve a fatality. If a fatal accident is
chosen at random, what is the probability the
accident involved a light truck?

14. Example 2
Events
C- Cars
T –Light truck
F –Fatal Accident
N- Not a Fatal Accident
Given
P(F|C) = 20/10000 and P(F|T) = 25/100000
P(T) = 0.4
In addition we know C and T are complementary events
P(C)=1-P(T)=0.6
Our goal is to compute the conditional probability of a Light truck
accident given that it is fatal P(T|F).
)TC(
c


15. Goal P(T|F)

Consider P(T|F)
Conditional probability
of a Light truck accident
given that it is fatal
Using conditional
probability formula
)F(P
)FT(P
)F|T(P


S=CT
C
T
F

16. P(T|F)-Method1
Consider P(T|F)
Conditional probability of a Light truck
accident given that it is fatal
How do we calculate?
Using conditional probability formula
)C(P)C|F(P)T(P)T|F(P
)T(P)T|F(P
)F(P
)FT(P
)F|T(P





).)(.().)(.(
).)(.(
600002040000250
40000250

=
= 0.4545

17. Tree Diagram- Method2
Vehicle
C
T
F C  F 0.6 0.0002 = .00012
F T  F 0.40.00025= 0.0001
N C  N 0.6 0.9998 = 0.59988
N T N 0.40.99975= .3999
0.6
0.4
0.9998
0.0002
0.00025
0.99975
Probability
Conditional
Probability
ProbabilityEvent

18. Tree Diagram- Method2
)FC(P)FT(P
)FT(P
)F(P
)FT(P
)F|T(P





).)(.().)(.(
).)(.(
600002040000250
40000250

=
= 0.4545

19. Partition
S
B
1 B
2
B
3
A
)()|()()|()()|()(
332211
BPBAPBPBAPBPBAPAP 

20. Law of Total Probability
))()()((
))((
)()(
21
21
n
n
BABABAP
BBBAP
SAPAP





Let the events B
1, B
2, , B
n partition the finite discrete sample
space S for an experiment and let A be an event defined on S.

21. Law of Total Probability






n
i
ii
nn
n
n
BPBAP
BPBAPBPBAPBPBAP
BAPBAPBAP
BABABAP
1
2211
21
21
)()|(
)()|()()|()()|(
)()()(
))()()((



.)()|()(
1



n
i
iiBPBAPAP

22. Bayes’ Theorem
Suppose that the events B
1, B
2, B
3, . . . , B
n
partition the sample space S for some
experiment and that A is an event defined on
S. For any integer, k, such that
we have
nk1





n
j
jj
kk
k
BPBAP
BPBAP
ABP
1
|
|
|

23. Focus on the Project
Recall
P(Y  T  C|S) will be used to calculate
P(S|Y  T  C)
P(Y  T  C|F) will be used to calculate
P(F|Y  T  C)

24. How can Bayes’ Theorem help us with the
decision on whether or not to attempt a loan work
out?
Partitions
1.Event S
2.Event F
Given
P(Y  T  C|S)
P(Y  T  C|F)
Need
P(S|Y  T  C)
P(F|Y  T  C)

25. Using Bayes Theorem
P(S|Y  T  C)  0.477
 
)536.0()021.0()464.0()022.0(
)464.0()022.0(
)()|()()|(
)()|(
|






FPFCTYPSPSCTYP
SPSCTYP
CTYSP
 
.
)536.0()021.0()464.0()022.0(
)536.0()021.0(
)()|()()|(
)()|(
|






FPFCTYPSPSCTYP
FPFCTYP
CTYFP
LOAN FOCUS EXCEL-BAYES
P(F|Y  T  C)  0.523

26. RECALL
Z is the random variable giving the amount of money,
in dollars, that Acadia Bank receives from a future
loan work out attempt to borrowers with the same
characteristics as Mr. Sanders, in normal times.
)523.0(000,250$)477.0(000,000,4$
)|(000,250$)|(000,000,4$
)000,250$(000,250$)000,000,4$(000,000,4$)(



CTYFPCTYSP
ZPZPZE
E(Z)  $2,040,000.

27. Decision
EXPECTED VALUE OF A WORKOUT=E (Z)  $2,040,000
FORECLOSURE VALUE- $2,100,000
RECALL
FORECLOSURE VALUE> EXPECTED VALUE OF A WORKOUT
DECISION
FORECLOSURE

28. Further Investigation I
let Y  be the event that a borrower has 6, 7, or 8 years of
experience in the business.
Using the range
Let Z be the random variable giving the amount of
money, in dollars, that Acadia Bank receives from a
future loan work out attempt to borrowers with Y

 and a
Bachelor’s Degree, in normal times. When all of the
calculations are redone, with Y

 replacing Y, we find that P(Y

  T  C|S)  0.073 and P(Y

  T  C|F)  0.050.
Former Bank
Years In
Business
Years In
Business
Education
Level
State Of
Economy
Loan Paid
Back
BR >=6 <=8 yes

29. Calculations
P(Y   T  C|S)  0.073
P(Y   T  C|F)  0.050
P(S|Y   T  C)  0.558
P(F|Y   T  C)  0.442
The expected value of Z is E(Z )  $2,341,000.
Since this is above the foreclosure value of
$2,100,000, a loan work out attempt is indicated.

30. Further Investigation II
 Let Y" be the event that a borrower has 5, 6,
7, 8, or 9 years of experience in the business
 Let Z" be the random variable giving the
amount of money, in dollars, that Acadia
Bank receives from a future loan work out
attempt to borrowers with 5, 6, 7, 8, or 9
years experience and a Bachelor's Degree, in
normal times. Redoing our work yields the
follow results.

31. Similarly can calculate E(Z  )
Make at a decision- Foreclose vs. Workout
Data indicates Loan work out

32. Close call for Acadia Bank loan officers
Based upon all of our calculations, we
recommend that Acadia Bank enter into a
work out arrangement with Mr. Sanders.